Xét : \(\left\{\begin{matrix}\left(x-8\right)^{2016}\ge0\\\sqrt{y-10}\ge0\end{matrix}\right.\)

=> Để \(\left(x-8\right)^{2016}+\sqrt{y-10}=0\)

Thì ( x- 8)²⁰¹⁶= \(\sqrt{y-10}\)= 0

\(\Rightarrow\left\{\begin{matrix}\left(x-8\right)^{2016}=0\\\sqrt{y-10}=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=8\\y=10\end{matrix}\right.\)

=> x+ y= 8+ 10= 18

Vậy x+ y= 18

Ta có 2 trường hợp:

Th1: (x-8)²⁰¹⁶ và \(\sqrt{y-10}\) là 2 số trài dấu.

Nhưng \(\left(x-8\right)^{2016}\ge0\) \(\forall x\)

\(\sqrt{y-10}\ge0\) \(\forall y\)

\(\Rightarrow\)(x-8)²⁰¹⁶ và \(\sqrt{y-10}\) ko thể trái dấu

Th2: \(\left(x-8\right)^{2016}=\sqrt{y-10}=0\)

\(\Rightarrow\left\{\begin{matrix}\left(x-8\right)=0\\y-10=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=8\\y=10\end{matrix}\right.\)

Vậy x+y=8+10=18

Vì \(\left(x-8\right)^{2016}\ge;\sqrt{y-10}\ge0\)

\(\Rightarrow\left(x-8\right)^{2016}+\sqrt{y-10}\ge0\)

Mà \(\left(x-8\right)^{2016}+\sqrt{y-10}=0\) \(\Rightarrow\left(x-8\right)^{2016}=0;\sqrt{y-10}=0\)

\(\Rightarrow x-8=0;y-10=0\)

\(\Rightarrow x=8;y=10\)

Ta có :(x - 8)^{2016 + \(\sqrt{y}-10\)} = 0

 ( x- 8 )²⁰¹⁶ >=0 ; \(\sqrt{y-10}>=0\)

=> ( x- 8 ) = 0 => x= 8

=> (y - 10 ) =0 => y = 10 

 => x+y = 8+10 

=> x+y = 18

1.

ĐKXĐ: $x\geq 1; y\geq 2; z\geq 3$

PT \(\Leftrightarrow x+y+z+8-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}=0\)

\(\Leftrightarrow [(x-1)-2\sqrt{x-1}+1]+[(y-2)-4\sqrt{y-2}+4]+[(z-3)-6\sqrt{z-3}+9]=0\)

\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{y-2}-2)^2+(\sqrt{z-3}-3)^2=0\)

\(\Rightarrow \sqrt{x-1}-1=\sqrt{y-2}-2=\sqrt{z-3}-3=0\)

\(\Leftrightarrow \left\{\begin{matrix} x=2\\ y=6\\ z=12\end{matrix}\right.\)

2.

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow \sqrt{x+1}=1-\sqrt{x}$

$\Rightarrow x+1=(1-\sqrt{x})^2=x+1-2\sqrt{x}$

$\Leftrightarrow 2\sqrt{x}=0$

$\Leftrightarrow x=0$

Thử lại thấy thỏa mãn 

Vậy $x=0$

3.

ĐKXĐ: $x\geq -1$

PT \(\Leftrightarrow (1+\sqrt{x^2+4033}).\frac{(x+2016)-(x+1)}{\sqrt{x+2016}+\sqrt{x+1}}=2015\)

\(\Leftrightarrow 1+\sqrt{x^2+4033}=\sqrt{x+2016}+\sqrt{x+1}\)

\(\Leftrightarrow (1+\sqrt{x^2+4033})^2=(\sqrt{x+2016}+\sqrt{x+1})^2\)

Áp dụng BĐT Bunhiacopxky:

\(\text{VP}\leq 2(x+2016+x+1)=4x+4034\)

\(\text{VP}=x^2+4034+2\sqrt{x^2+4033}\geq x^2+4034+2\sqrt{4033}>x^2+4034+5\)

Mà: $x^2+4034+5-(4x+4034)=(x-2)^2+1> 0$

$\Rightarrow x^2+4034+5> 4x+4034$

$\Rightarrow \text{VP}> \text{VT}$

Do đó pt vô nghiệm.

bạn tách từng câu ra mik suy nghĩ từng câu

Bạn thêm điều kiện x,y,z lớn hơn 0 nhé :)

Từ giả thiết ta suy ra : \(a^2=b+4032\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+4032\)

\(\Rightarrow xy+yz+zx=2016\)thay vào :

\(x\sqrt{\frac{\left(2016+y^2\right)\left(2016+z^2\right)}{2016+x^2}}=x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{x^2+xy+yz+zx}}\)

\(=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(z+y\right)\left(z+x\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}=x\left|y+z\right|=xy+xz\)vì x,y,z > 0

Tương tự : \(y\sqrt{\frac{\left(2016+z^2\right)\left(2016+x^2\right)}{2016+y^2}}=xy+zy\)

\(z\sqrt{\frac{\left(2016+x^2\right)\left(2016+y^2\right)}{2016+z^2}}=zx+zy\)

Suy ra \(P=2\left(xy+yz+zx\right)=2.2016=4032\)